Question

Give reason for the statement, '\lbrack Ni(CN)_{4}\rbrack^{2 -}'is diamagnetic while $\lbrack NiCl_{4}\rbrack^{2 -}$is paramagnetic in nature.

• A. In $\lbrack NiCl_{4}\rbrack^{2 -}$on unpaired electrons are present while in $\lbrack Ni(CN)_{4}\rbrack^{2 -}$ two unpaired electrons are present.
• B. In $\lbrack Ni(CN)_{4}\rbrack^{2 -}$on unpaired electrons are present while in $\left\lbrack NiCl_{4} \right\rbrack^{2 -}$two unpaired electons are present.
• C. $\lbrack NiCl_{4}\rbrack^{2 -}$shows $dsp^{2}$hybridizations hences it is paramagnetic.
• D. $\lbrack Ni(CN)_{4}\rbrack^{2 -}$shows $sp^{3}$hybridization hence it is diamagnetic.

Answer 1

Answer: (B)

In $\lbrack Ni(CN)_{4}\rbrack^{2 -}$on unpaired electrons are present while in $\left\lbrack NiCl_{4} \right\rbrack^{2 -}$two unpaired electons are present.

Answer 2

$\lbrack Ni(CN)_{4}\rbrack^{2 -}$there is no unpaired electron because $CN^{-}$is a strong filed ligand. Therefore it is diamagnetic in nature.

In $\lbrack NiCl_{4}\rbrack^{2 -}$there are two unpaired electrons because $Cl^{-}$is a weak field ligand. Therefore, it is paramagnetic in nature.